solutions of 1+x+x2+x3=y2

This article shows that the only solutions in integers to the equation

1+x+x2+x3=y2

are the obvious trivial solutions x=0,±1 together with x=7,y=20. This result was known to Fermat.

First, note that the equation is (1+x)⁢(1+x2)=y2, so we have immediately that x≥-1. So, noting the solutions for x=0,±1, assume in what follows that x>1.

Let d=gcd⁡(1+x,1+x2). Then x≡-1(modd), so that 1+x2≡2(modd). But d∣1+x2 so that 2≡0(modd) and d is either 1 or 2. If d=1, so that 1+x and 1+x2 are coprime, then 1+x and 1+x2 must both be squares. But 1+x2 is not a square for x>1. Thus the gcd of 1+x and 1+x2 is 2, so each must be twice a square, say

Recall that if (a,b,c) is a primitive Pythagorean triple, then precisely one of a and b is even, and we can choose coprime integers p,q such that a=p2-q2,b=2⁢p⁢q,c=p2+q2 or a=2⁢p⁢q,b=p2-q2,c=p2+q2 depending on the parity of a.

Suppose first that r is odd. Then

r2=p2-q2

r2-1=2⁢p⁢q

s=p2+q2

Then 1=r2-(r2-1)=(p-q)2-2⁢q2. Now, note that p-q must be a square, say p-q=t2, since gcd⁡(p-q,p+q)=1 and (p-q)⁢(p+q) is a square. Then

t4=(p-q)2=2⁢q2+1=(q2+1)2-q4

so that

t4+q4=(q2+1)2

But we know that the sum of two fourth powers can be a square (http://planetmath.org/ExampleOfFermatsLastTheorem) only for the trivial case where all are zero. So r cannot be odd.

So suppose that r is even. Then

r2=2⁢p⁢q

r2-1=p2-q2

s=p2+q2

From the second of these formulas, we see that p must be even (consider both sides (mod4)), say p=2⁢t2. Now,

Thus one of u,u+1 is a fourth power and the other is twice a fourth power, say b4 and 2⁢c4.

Now,

u=b4⇒u+1=2⁢c4⇒b4-2⁢c4=-1

u=2⁢c4⇒u+1=b4⇒b4-2⁢c4=1

so that b4-2⁢c4=±1.

If b4-2⁢c4=1, then

((c2)2+1)2=(c2)4+b4

and again we have a square being the sum of two fourth powers. So this case is impossible.

If b4-2⁢c4=-1, write e=c2, then (e2-1)2=e4-b4. It follows (see here (http://planetmath.org/X4Y4z2HasNoSolutionsInPositiveIntegers)) that either b=0 or e2-1=0. If b=0, we get the impossibility e4-2⁢e2+1=e4, while if (e2-1)2=0, then e4=b4,e=±1 and so b=±1. Then p+q-12=b4=1, so that p+q=3. Thus r2=4, so r=2 and, finally, 1+x=2⁢r2=8, so that x=7.